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Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.

Let  the required numbers be a and b.

Given, 14 is the mean proportional between a and b.

\[\begin{array}{*{35}{l}}

a:\text{ }14\text{ }=\text{ }14:\text{ }b  \\

ab\text{ }=\text{ }196  \\

a\text{ }=\text{ }196/b\text{ }\ldots .\text{ }\left( 1 \right)  \\

\end{array}\]

Also, given, third proportional to a and b is 112.

\[\begin{array}{*{35}{l}}

a:\text{ }b\text{ }=\text{ }b:\text{ }112  \\

{{b}^{2}}~=\text{ }112a\text{ }\ldots .\text{ }\left( 2 \right)  \\

\end{array}\]

Using (1), we have:

\[\begin{array}{*{35}{l}}

{{b}^{2}}~=\text{ }112\text{ }\times \text{ }\left( 196/b \right)  \\

{{b}^{3}}~=\text{ }{{14}^{3}}~x\text{ }{{2}^{3}}  \\

b\text{ }=\text{ }28  \\

\end{array}\]

From (1),

a = 196/ 28 = 7

Therefore, the two numbers are 7 and 28.